# arctan integral

• Mar 24th 2010, 06:16 PM
coscos
arctan integral
Hi! I just read http://www.mathhelpforum.com/math-he...808-post2.html and the bottom line says
Quote:

Originally Posted by jmcq
The long and short of it is when you have an integral of the form $\int\frac{dx}{{a^2+x^2}}$ you get an result of the form: $\frac{1}{a} \arctan \left(\frac{x}{a}\right)+C$

It's a good one to memorize.

I am trying to do a similar thing but I'm ending up getting the same (wrong) answer all the time.

$\int \frac{dx}{(x+2)^2+4}\\=$
$\int \frac{dx}{(\frac{1}{4} \cdot4)\cdot (x+2)^2+4}\\=$
$\frac{1}{4}\int \frac{dx}{(\frac{1}{2})^2(x+2)^2+1}\\=$
$\frac{1}{4}\int \frac{dx}{(\frac{x+2}{2})^{2}+1}=\frac{1}{4}\left[ arctan(\frac{x+2}{2})+C \right]$

I don't know where the error is, can anyone please show me? It would really appreciated! Please just ask if I've expressed myself unclear!
• Mar 24th 2010, 06:28 PM
skeeter
$\int \frac{1}{(x+2)^2 + 4} \, dx$

$\int \frac{1}{4\left[\left(\frac{x+2}{2}\right)^2 + 1\right]} \, dx$

$\frac{2}{4} \int \frac{\frac{1}{2}}{\left(\frac{x+2}{2}\right)^2 + 1} \, dx$

$\frac{1}{2} \arctan\left(\frac{x+2}{2}\right) + C$
• Mar 24th 2010, 06:35 PM
coscos
Quote:

Originally Posted by skeeter
$\int \frac{1}{(x+2)^2 + 4} \, dx$

$\int \frac{1}{4\left[\left(\frac{x+2}{2}\right)^2 + 1\right]} \, dx$

$\frac{2}{4} \int \frac{\frac{1}{2}}{\left(\frac{x+2}{2}\right)^2 + 1} \, dx$

$\frac{1}{2} \arctan\left(\frac{x+2}{2}\right) + C$

Oh : D Of course! Thank you very much!